Question

Potential energy of a particle at position x is given by u=x^2-5x

Answer 1

Answer:

Potential is minimum at x=2.5

Minimum Potential = -6.25

Explanation:

Given that,

$u = x^{2}-5x$....(I)

On differentiate of u

$\dfrac{du}{dx} = 2x-5$

Now,

$\dfrac{du}{dx} = 0$

Then, $2x-5 = 0$

$2x = 5$

$x = \dfrac{5}{2}$

$x = 2.5$

Now, double derivative is

$\dfrac{d^{2}u}{dx^{2}} = 2$

At x = 2.5, double derivative is positive.

So, put the value of x in equation (I)

$u = (2.5)^{2}-5\times2.5$

$u = -6.25$

Hence, potential is minimum at x=2.5 Minimum Potential = -6.25

Answer 2

"x=2.5

As we know that a particle is said to be in equilibrium if net force acting on the particle is zero .

we can also go through this formula

${ F }_{ x }\quad =\quad \frac { du }{ dx }$

For Force=0 we have to make $\frac { dU }{ dX } \quad =\quad 0$

now put the value in first derivative i.e 2X-5=0 .

x=2.5 ans"